If ##Y = \exp(X)##, then the cdf of ##Y## is ##F_Y(\alpha) = P(\exp(X) \leq \alpha)##. As ##\log## is strictly monotonically increasing, this is equivalent to ##P(X \leq \log(\alpha))##, which is ##F_X(\log(\alpha))##, the cdf of ##X## evaluated at ##\log(\alpha)##.
Then you can find the pdf of ##Y## by differentiating ##F_Y(\alpha) = F_X(\log\alpha))## to obtain ##f_Y(\alpha) = f_X(\log(\alpha)) / \alpha##. This is valid for all positive ##\alpha##; if ##\alpha## is negative then the pdf is zero since ##\exp## is never negative.
Proceeding from here should be straightforward as there is a closed form expression for ##f_X##.